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Chemistry

2 answers:

astraxan [27]4 years ago

7 0

i hope this helped!

inysia [295]4 years ago

6 0

None of the above are correct.

Albert Einstein supported the use of radioactive elements for medical diagnostic purposes.

I hope this helps, good luck!

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2K + 2HBr → 2 KBr + H2

Inessa [10]

Answer:

$\large \boxed{\text{0.0503 g}}$

Explanation:

The limiting reactant is the reactant that gives the smaller amount of product.

Assemble all the data in one place, with molar masses above the formulas and masses below them.

M_r: 39.10 80.41 2.016

2K + 2HBr ⟶ 2KBr + H₂

m/g: 5.5 4.04

a) Limiting reactant

(i) Calculate the moles of each reactant

$\text{Moles of K} = \text{5.5 g} \times \dfrac{\text{1 mol}}{\text{31.10 g}} = \text{0.141 mol K}\\\\\text{Moles of HBr} = \text{4.04 g} \times \dfrac{\text{1 mol}}{\text{80.91 g}} = \text{0.049 93 mol HBr}$

(ii) Calculate the moles of H₂ we can obtain from each reactant.

From K:

The molar ratio of H₂:K is 1:2.

$\text{Moles of H}_{2} = \text{0.141 mol K} \times \dfrac{\text{1 mol H}_{2}}{\text{1 mol K}} = \text{0.0703 mol H}_{2}$

From HBr:

The molar ratio of H₂:HBr is 3:2.

$\text{Moles of H}_{2} = \text{0.049.93 mol HBr } \times \dfrac{\text{1 mol H}_{2}}{\text{1 mol HBr}} = \text{0.024 97 mol H}_{2}$

(iii) Identify the limiting reactant

HBr is the limiting reactant because it gives the smaller amount of NH₃.

b) Excess reactant

The excess reactant is K.

c) Mass of H₂

$\text{Mass of H}_{2} = \text{0.024 97 mol H}_{2} \times \dfrac{\text{2.016 g H}_{2}}{\text{1 mol H}_{2}} = \textbf{0.0503 g H}_{2}\\ \text{The mass of hydrogen is $\large \boxed{\textbf{0.0503 g}}$ }$